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When dealing with a right angled triangle trigonometric functions can be defined by :-
cos x = adjacent/hypotenuse
sin x = opposite/hypotenuse
Therefore cos x/sin x = adj/hyp ÷ opp/hyp = adj/hyp x hyp/opp = adj/opp
The tangent of an angle is given by the formula : opposite/adjacent
The tangent of the complementary angle is therefore : adjacent/opposite
Then cos x/sin x = tan(90-x)
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How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
Find the critical numbers of sinx plus cosx?
f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
What is the derivative of the square root of 2 raised to the power of cos x?
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
What is the derivative of sin x minus cos x?
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
What is the derivative of -cosx?
-sinx
How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
What is the derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
Find the critical numbers of sinx plus cosx?
f(x)=sinx+cosx take the derivative f'(x)=cosx-sinx critical number when x=pi/4
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
Find the derivative of sinxcosx?
(cosx)^2-(sinx)^2
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
What is the derivative of cos x?
d/dx(-cosx)=--sinx=sinx
What is the derivative of the square root of 2 raised to the power of cos x?
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
What is the anti-derivative of 4cosx divided by sinx to the power of 2?
2
What is the derivative of 7 to the power of cosx?
derivative (7cosx) = -ln(7) 7cosx sinx dx
